# Chapter 2 Section 2 Solving a System of Linear Equations II.

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Chapter 2 Section 2 Solving a System of Linear Equations II

Recall from Algebra A system of linear equations has one of the following results: 1.A unique solution 2.An infinite number of solutions 3.No solution

A Unique Solution That each variable is equated to a number. The examples and exercises in Chapter 2 Section 1 are systems of equations that have a unique solution.

Exercise 11 (page 76) x + 2y = 5 3x – y = 1 – x + 3y = 5 1 2 5 3 – 1 1 – 1 3 5 rref 1 2 5 3 – 1 1 – 1 3 5 1 0 1 0 1 2 0 0 0,, 1x + 0y = 1 0x + 1y = 2 0x + 0y = 0 x = 1 y = 2 0 = 0 x = 1 and y = 2 (A unique solution)

An Infinite Number of Solutions If there are (1) no inconsistent equations (we’ll talk about these in the case of no solutions) and (2) there is/are variable(s) that are not assigned a value, then there are an infinite number of solutions The trick here is to get the resulting rref matrix and convert the rref matrix back into equation notation.

Exercise 15 (page 76) x + y + z = – 1 2x + 3y + 2z = 3 2x + y + 2z = – 7 1 1 1 – 1 2 3 2 3 2 1 2 – 7 rref, 1x + 0y + 1z = – 6 0x + 1y + 0z = 5 0x + 0y + 0z = 0 x + z = – 6 y = 5 0 = 0 1 1 1 – 1 2 3 2 3 2 1 2 – 7 1 0 1 – 6 0 1 0 5 0 0 0 0

Exercise 15 continued x + z = – 6 y = 5 0 = 0 x = – z – 6 y = 5 z = any value Answer: x = – z – 6, y = 5, and z = any value (This is a case of an infinite number of solutions) (Note that 0 = 0 is a true statement)

Exercise 19 (page 76) x + y – 2z + 2w = 5 2x + 1y – 4z + w = 5 3x + 4y – 6z + 9w = 20 4x + 4y – 8z + 8w = 20 rref, 1x + 0y – 2z – 1w = 0 0x + 1y + 0z + 3w = 5 0x + 0y + 0z + 0w = 0 x – 2z – w = 0 y + 3w = 5 0 = 0 1 0 – 2 – 1 0 0 1 0 3 5 0 0 0 0 0 1 1 – 2 2 5 2 1 – 4 1 5 3 4 – 6 9 20 4 4 – 8 8 20 1 1 – 2 2 5 2 1 – 4 1 5 3 4 – 6 9 20 4 4 – 8 8 20

Exercise 19 continued x = 2z + w y = – 3w + 5 z = any value w = any value Answer: x = 2z – w, y = – 3w + 5, z = any value, and w = any value (This is a case of an infinite number of solutions) x – 2z – w = 0 y + 3w = 5 0 = 0 (Note that 0 = 0 are true statements)

No Solution The trick here is to look for an inconsistent row. (i.e. 0 = 1 row)

Exercise 39 (page 77) 4x – 3y + 2z = 3 – 7x + 5y = 2 – 10x + 7y + 2z = 4 4 –3 2 3 – 7 5 0 2 –10 7 2 4 rref, 1x + 0y – 10z = 0 0x + 1y – 14z = 0 0x + 0y + 0z = 1 x – 10z = 0 y – 14z = 0 0 = 1 1 0 – 10 0 0 1 – 14 0 0 0 0 1 4 –3 2 3 – 7 5 0 2 –10 7 2 4

Exercise 39 continued Answer: There is no solution to the system of equations Note that 0 = 1 is an inconsistent equation (i.e. a false statement), thus the system has no solution. x – 10z = 0 y – 14z = 0 0 = 1

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